Evaluating $\lim\limits_{n \to \infty} \left(1+{3 \over n}\right)^n$ 
$$\lim_{n \to \infty} \left(1+{3 \over n}\right)^n$$

What are the general rules for limit of this kind, like $\lim_{n \to \infty} \left(1+{\alpha \over n}\right)^n$ or $\lim_{n \to \infty} \left(1+{\alpha \over \beta n}\right)^n$ 
And how can I solve this? 
 A: Notice, we know that $$\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n=e$$
General rule: let $$\frac{\alpha}{\beta n}=\frac{1}{t}\implies n=\frac{\alpha }{\beta }t$$ hence, we get $$\lim_{n\to \infty}\left(1+\frac{\alpha}{\beta n}\right)^n=\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{\frac{\alpha }{\beta }t}$$ $$=\left(\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{t}\right)^{\frac{\alpha }{\beta }}$$$$=\large{e^{\frac{\alpha }{\beta }}}$$
A: $$\text{Let }L = \lim_{n\to\infty} \left(1 + \frac cn\right)^n$$
$$\log L = \lim_{n\to\infty} n\log\left(1 + \frac cn\right) = \frac{\log\left(1 + \frac cn\right)}{\frac 1n} = \frac{1}{1 + \frac cn}\frac{\frac{c}{n^2}}{\frac 1{n^2}} = c$$
$$\log L = c \implies L = e^c$$
A: Here is one way of approaching this.
Note that $|\ln (1+x) - x| \le {1 \over 2} x^2$ if $|x|<1$.
Suppose $x_n \to x$. Choose $n$ large enough so that $|x_n| < n$, then
$(1+{x_n \over n})^n = e^{n \ln (1+{x_n \over n}) }$.
Since $ |n\ln (1+{x_n \over n}) - x_n| \le {1 \over n } x_n^2$, we see that ${n \ln (1+{x_n \over n})  \to x} $, and so
$(1+{x_n \over n})^n \to e^x$.
A: $$\begin{align}
\lim_{n \to \infty} \left(1+\frac{\alpha}{\beta n}\right)^n
 &= \lim_{n \to \infty} \left(1+\frac1{\frac{\beta n}{\alpha}}\right)^{\dfrac{\beta n}{\alpha}\cdot\dfrac{\alpha}{\beta}} \\[2 ex]
 &= \lim_{\frac{\beta n}{\alpha} \to \infty} \left[\left(1+\frac1{\frac{\beta n}{\alpha}}\right)^{\dfrac{\beta n}{\alpha}}\right]^{\dfrac{\alpha}{\beta}} \\[2 ex]
 &= \left[\lim_{\frac{\beta n}{\alpha} \to \infty} \left(1+\frac1{\frac{\beta n}{\alpha}}\right)^{\dfrac{\beta n}{\alpha}}\right]^{\dfrac{\alpha}{\beta}} \\[2 ex]
 &= \left[\lim_{m \to \infty} \left(1+\frac1{m}\right)^{m}\right]^{\dfrac{\alpha}{\beta}} \\[2 ex]
 &= e^{\frac{\alpha}{\beta}}
\end{align}$$
The main tricky part here is taking the constant exponent outside the limit, but an obvious theorem says this is permissible if the limit without the exponent exists, and it clearly does here. Replacing $n$ with $m$ here is less tricky but is still clearly permitted here. Ask if you need full justification of either of those "tricks."
A: $$\lim_{n\to\infty}\left(1+\frac{3}{n}\right)^n=\lim_{n\to\infty}\exp\left(\ln\left(\left(1+\frac{3}{n}\right)^n\right)\right)=\lim_{n\to\infty}\exp\left(n\ln\left(1+\frac{3}{n}\right)\right)=$$
$$\exp\left(\lim_{n\to\infty}n\ln\left(1+\frac{3}{n}\right)\right)=\exp\left(\lim_{n\to\infty}\frac{\ln\left(1+\frac{3}{n}\right)}{\frac{1}{n}}\right)=$$
$$\exp\left(\lim_{n\to\infty}\frac{\frac{\text{d}}{\text{d}n}\ln\left(1+\frac{3}{n}\right)}{\frac{\text{d}}{\text{d}n}\left(\frac{1}{n}\right)}\right)=\exp\left(\lim_{n\to\infty}\frac{-\frac{3}{n^2\left(1+\frac{3}{n}\right)}}{-\frac{1}{n^2}}\right)=$$
$$\exp\left(\lim_{n\to\infty}\frac{3n}{n+3}\right)=\exp\left(\lim_{n\to\infty}\frac{3}{1+\frac{3}{n}}\right)=\exp\left(\frac{3}{1+0}\right)=\exp(3)=e^3$$


In General:
$$\lim_{n\to\infty}\left(1+\frac{b}{cn}\right)^n=$$
$$\lim_{n\to\infty}\exp\left(\ln\left(\left(1+\frac{b}{cn}\right)^n\right)\right)=$$
$$\lim_{n\to\infty}\exp\left(n\ln\left(1+\frac{b}{cn}\right)\right)=$$
$$\exp\left(\lim_{n\to\infty}n\ln\left(1+\frac{b}{cn}\right)\right)=$$
$$\exp\left(\lim_{n\to\infty}\frac{\ln\left(1+\frac{b}{cn}\right)}{\frac{1}{n}}\right)=$$
$$\exp\left(\lim_{n\to\infty}\frac{\frac{\text{d}}{\text{d}n}\ln\left(1+\frac{b}{cn}\right)}{\frac{\text{d}}{\text{d}n}\left(\frac{1}{n}\right)}\right)=$$
$$\exp\left(\lim_{n\to\infty}\frac{-\frac{b}{cn^2+bn}}{-\frac{1}{n^2}}\right)=$$
$$\exp\left(\lim_{n\to\infty}\frac{bn}{cn+b}\right)=$$
$$\exp\left(\lim_{n\to\infty}\frac{b}{c+\frac{b}{n}}\right)=$$
$$\exp\left(\frac{b}{c+0}\right)=$$
$$\exp\left(\frac{b}{c}\right)=e^{\frac{b}{c}}$$
